Exercise on three-circles theorems

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Math 618
Theory of Functions of a Complex Variable II
Spring 2015
Exercise on three-circles theorems
In this assignment, you will apply subharmonic functions to prove Hadamard’s
three-circles theorem (which is one of the topics on the official syllabus for the
qualifying examination in complex analysis).
1. Suppose 𝑒 is a subharmonic function in an annulus with inner radius π‘Ž and
outer radius 𝑏: namely, { 𝑧 ∈ β„‚ ∢ π‘Ž < |𝑧| < 𝑏 }. Let π‘š(π‘Ÿ) denote the
maximum of 𝑒(𝑧) when |𝑧| = π‘Ÿ. Prove that if π‘Ž < π‘Ÿ1 < π‘Ÿ < π‘Ÿ2 < 𝑏, then
π‘š(π‘Ÿ) ≤
log π‘Ÿ − log π‘Ÿ1
log π‘Ÿ2 − log π‘Ÿ
π‘š(π‘Ÿ1 ) +
π‘š(π‘Ÿ2 ).
log π‘Ÿ2 − log π‘Ÿ1
log π‘Ÿ2 − log π‘Ÿ1
Sometimes this inequality is described by saying that π‘š(π‘Ÿ) is “a convex function of log π‘Ÿ.” An equivalent formulation of the inequality is that
log
π‘š(π‘Ÿ) ≤
log
π‘Ÿ2
π‘Ÿ
π‘Ÿ2
π‘Ÿ1
π‘š(π‘Ÿ1 ) +
log π‘Ÿπ‘Ÿ
1
log
π‘Ÿ2
π‘Ÿ1
π‘š(π‘Ÿ2 ).
Suggestion: Use a harmonic comparison function of the form 𝐴 + 𝐡 log |𝑧|.
2. Deduce that if 𝑓 is a holomorphic function in the annulus, and 𝑀(π‘Ÿ) denotes
the maximum of |𝑓 (𝑧)| when |𝑧| = π‘Ÿ, then
πœ†
1−πœ†
𝑀(π‘Ÿ) ≤ 𝑀(π‘Ÿ1 ) 𝑀(π‘Ÿ2 )
log
,
where π‘Ž < π‘Ÿ1 < π‘Ÿ < π‘Ÿ2 < 𝑏, and πœ† =
log
π‘Ÿ2
π‘Ÿ
π‘Ÿ2
π‘Ÿ1
.
This statement is Hadamard’s three-circles theorem (Theorem 23.0.3 on
page 387 of the textbook).
3. Suppose in Part 1 that 𝑒 is bounded above, and π‘Ž = 0. Let π‘Ÿ1 approach 0 to
deduce that π‘š(π‘Ÿ) is an increasing function of π‘Ÿ. Extend 𝑒 to the whole disk
{ 𝑧 ∈ β„‚ ∢ |𝑧| < 𝑏 } by declaring 𝑒(0) to be limπ‘Ÿ→0+ π‘š(π‘Ÿ). Prove that the
extended function 𝑒 is subharmonic on the whole disk.
This statement is a removable singularity theorem for subharmonic functions
that are bounded above on a punctured disk.
April 2, 2015
Page 1 of 1
Dr. Boas
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